## zzr0ck3r one year ago Good methods for finding eigen values? $$M = \begin{array}{cc} 0&1&1&1&1& \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&1\\ 1&1&1&1&0 \end{array}$$

1. zzr0ck3r

@freckles where you go ;(

2. UsukiDoll

$\det(A- \lambda I) =0$ and then use cofactor expansion like crazy

3. zzr0ck3r

I said good methods :)

4. UsukiDoll

=(

5. UsukiDoll

row operations first? idk that may destroy the determinant...

6. freckles

lol well ... I was kind of trying to see if could find a pattern but I didn't see a pattern like you know M_n is a n by n matrix with 0's on the diagonal and 1's everywhere else but it was totally not a good method

7. freckles

and not just a bad method but it didn't seem to work like that

8. UsukiDoll

does this whole matrix break one of the rules for row operations? Don't we need a 1 at the top left corner of the first row?

9. zzr0ck3r

word

10. UsukiDoll

row swaps... so it does not break the first rule

11. UsukiDoll

I would switch the last one with the first one r_1<->r_5

12. UsukiDoll

swap r_2 <-> r_4 , then r_3 <-> r_4 ?

13. freckles

actually I do see a pattern

14. freckles

so there must be some kinda of shortcut for M_n where M_n is a n by n matrix with 0's a long the diagonal and 1's elsewhere

15. freckles

I want to make a conjecture $\text{ egienvalues }(M_n)=n-1,-1 \text{ (multiplicity } (n-1) \text{ )}$

16. freckles

Now I know this isn't a method but it might help to come up with something from this if it is even true

17. freckles

like we know the trace of M is suppose to be the sum of the eigenvalues

18. freckles

$0=\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5$

19. freckles

I don't know a quick way to find the determinant but the product of the eigenvalues equals the determinant

20. zzr0ck3r

ill play with this too

21. freckles

another cute thing $M_n=[1]_n-I$ where 1_n is a n by n matrix only made of 1's and I is the identity matrix don't know if this helps yet

22. freckles

you know I'm not totally sure yet but it seems to be: in this case anyways $\text{ eigenvalues(}M_n\text{)}=\text{ egivenvalues(}[1]_n\text{)}-\text{ egienvalues(}I\text{)}$

23. freckles

the eigenvalues of I are 1 (n amount of times) and the eigenvalues of [1]_n are n, 0 (n-1 amount of times)